When does replacement (accidentally) hold in amenable sets?

Question

A set $M$ is called amenable if it is transitive and satisfies the following conditions:

1. For all $x,y\in M$, $\{x,y\}\in M$
2. For all $x\in M$, $\bigcup x \in M$
3. $\omega \in M$
4. For all $x,y \in M$, $x\times y \in M$
5. ($\Sigma_0$ comprehension) Whenever $\Phi$ is a $\Sigma_0$ formula of one free variable with parameters from $M$, then for all $x\in M$, $\{z\in x | \Phi(z)\}\in M$

Although the definition of an amenable set does not include replacement, some very limited amount of replacement follows from the axioms given. For example, for all $x,y\in M$, it must be that $\{\{z,w\}|z\in x, w\in y\}\in M$ and $\{\{z\}|z\in x\}\in M$. So just how limited is the replacement in amenable sets? In particular,

If $M$ is an amenable set and $x\in M$, does it follow that $\{\bigcup z | z \in x\} \in M$?

Answer 1

I think the following is a counterexample to your specific question. Let AH be the set of those $x$ such that (1) each element of $TC\{x\}$ has cardinality at most $\aleph_\omega$ and (2) all but finitely many elements of $TC\{x\}$have cardinality strictly smaller than $\aleph_\omega$. (By $TC\{x\}$, I mean the transitive closure of the singleton, so it contains $x$, all its members, all their members, etc.) If $y\in x$ then $TC\{y\}$ is a subset of $TC\{x\}$, so AH is transitive. AH contains $\omega$ and is easily seen to be closed under pairing, union, and (binary) Cartesian product. Furthermore, it satisfies not only $\Sigma_0$-comprehension but full comprehension, because if $y$ is a subset of $x$ then $TC\{y\}$ is a subset of $TC\{x\}\cup\{y\}$. So AH is amenable.

For each natural number $n$, let $A_n=\{\{n\}\times\aleph_k:k\in\omega\}$, and notice that AH contains not only each of the $A_n$'s but also the set $X$ consisting of all the $A_n$'s. The union of any particular $A_n$ is $\{n\}\times\aleph_\omega$, which is in AH and has cardinality $\aleph_\omega$, but the set of all these unions is not in AH because it has these infinitely many elements of size $\aleph_\omega$. Summary: $X$ is in AH but $\{\bigcup z:z\in X\}$ is not.

Comment: If one modifies the definition of AH by requiring all elements of $TC\{x\}$ to have size strictly below $\aleph_\omega$, one gets the standard example of a model of all the ZFC axioms except the axiom of union. By allowing, in the definition of AH, finitely many exceptions of size $\aleph_\omega$ one revives the axiom of union and in particular one lets each of the sets $\bigcup A_n$ into AH (but just barely) but not the collection of all of them.

Answer 2

I agree with Andreas Blass's solution. The problem, or difficulty, with the definition of amenable is highlighted with this example: $\Sigma_0$-comprehension in this questioner's scheme is not really adequate.

For this reason it is sometimes replaced with $\Sigma_0$- (or `rudimentary') -closure for $\Delta_0$ formulae $\varphi$: $$\forall x \exists w \forall \vec v \in x \exists t\in w \forall u (u \in t \leftrightarrow u \in x \wedge \varphi[u, \vec v]) $$

This is more useful, implies $\Sigma_0$-Comprehension, and rules out the undesirable effect of the example.